Question

Let `f(x) = x^3` and compute the Riemann sum of f over the interval [2, 3], n=2 intervals using midpoints?

Answer 1

Representing the interval `[2,3]` as `[a,b]`, let `\Delta x=\frac{b-a}{n}=(3-2)/2=0.5`, `x_{0}=a=2`, `x_{1}=x_{0}+\Delta x=2+0.5=2.5`, and `x_{2}=x_{1}+\Delta x=2.5+0.5=3=b`.

Now let `x_{1}^{\star}=2.25`, the midpoint of `[2,2.5]` and `x_{2}^{\star}=2.75`, the midpoint of `[2.5,3]`.

The midpoint Riemann sum approximation in this situation is then

`f(x_{1}^{\star})*\Delta x+f(x_{2}^{\star})*\Delta x=(f(2.25)+f(2.75))*0.5`

`=(2.25^3+2.75^3)*0.5=(11.390625+20.796875)*0.5=16.09375`.

Compare this with the exact answer:

`\int_{2}^{3}x^{3}\ dx=x^{4}/4 |_{x=2}^{x=3}=\frac{81-16}{4}=65/4=16.25`